合成関数の差分の証明

$\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}=\frac{g(x+h)-g\left(x\right)}{(x+h)-x}$
$=\frac{g(x+h)-g\left(x\right)}{h}$
$\text{から変形して}$
$g(x+h)=g\left(x\right)+\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h\text{…①}$

$\text{差分(商)の公式に①を代入}$
$\frac{{∆}_{h}f\left(g\left(x\right)\right)}{{∆}_{h}x}=\frac{f\left(g(x+h)\right)-f\left(g\left(x\right)\right)}{(x+h)-x}$
$=\frac{f(g\left(x\right)+\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h)-f\left(g\left(x\right)\right)}{h}$
$=\frac{f(g\left(x\right)+\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h)-f\left(g\left(x\right)\right)}{h}\frac{\left(\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}\right)}{\left(\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}\right)}$
$=\frac{f(g\left(x\right)+\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h)-f\left(g\left(x\right)\right)}{\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}$
$=\frac{f(g\left(x\right)+\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h)-f\left(g\left(x\right)\right)}{(g\left(x\right)+\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}h)-g\left(x\right)}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}$
$=\frac{{∆}_{(h\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x})}f\left(g\left(x\right)\right)}{{∆}_{(h\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x})}g\left(x\right)}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}$

$\text{よって}$
$\frac{{∆}_{h}f\left(g\left(x\right)\right)}{{∆}_{h}x}=\frac{{∆}_{(h\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x})}f\left(g\left(x\right)\right)}{{∆}_{(h\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x})}g\left(x\right)}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}$

$\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}=\frac{g\left(x\right)-g(x-h)}{x-(x-h)}$
$=\frac{g\left(x\right)-g(x-h)}{h}$
$\text{から変形して}$
$g(x+h)=g\left(x\right)-h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}\text{…①}$

$\text{差分の公式に①を代入}$
$\frac{{\nabla }_{h}f\left(g\left(x\right)\right)}{{\nabla }_{h}x}=\frac{f\left(g\left(x\right)\right)-f\left(g(x-h)\right)}{x-(x-h)}$
$=\frac{f\left(g\left(x\right)\right)-f(g\left(x\right)-\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}h)}{h}$
$=\frac{f\left(g\left(x\right)\right)-f(g\left(x\right)-\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}h)}{h}\frac{\left(\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}\right)}{\left(\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}\right)}$
$=\frac{f\left(g\left(x\right)\right)-f(g\left(x\right)-\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}h)}{\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}h}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}$
$=\frac{f\left(g\left(x\right)\right)-f(g\left(x\right)-\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}h)}{g\left(x\right)-(g\left(x\right)-\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}h)}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}$
$=\frac{{\nabla }_{(h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x})}f\left(g\left(x\right)\right)}{{\nabla }_{(h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x})}g\left(x\right)}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}$

$\text{よって}$
$\frac{{\nabla }_{h}f\left(g\left(x\right)\right)}{{\nabla }_{h}x}=\frac{{\nabla }_{(h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x})}f\left(g\left(x\right)\right)}{{\nabla }_{(h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x})}g\left(x\right)}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}$

$M_{x}g\left(x\right)=\frac{g(x+\frac{h}{2})-g(x-\frac{h}{2})}{2}$
$=\frac{2g(x+\frac{h}{2})-g(x+\frac{h}{2})+g(x-\frac{h}{2})}{2}$
$=g(x+\frac{h}{2})-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$
$\text{式変形して}$
$g(x+\frac{h}{2})=M_{x}g\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}\text{…①}$

$M_{x}g\left(x\right)=\frac{g(x+\frac{h}{2})-g(x-\frac{h}{2})}{2}$
$=\frac{g(x+\frac{h}{2})-g(x-\frac{h}{2})+2g(x-\frac{h}{2})}{2}$
$=g(x-\frac{h}{2})+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$
$\text{式変形して}$
$g(x-\frac{h}{2})=M_{x}g\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}\text{…②}$

$\text{差分の公式に①と②を代入}$
$\frac{\delta f\left(g\left(x\right)\right)}{\delta x}=\frac{f\left(g(x+\frac{h}{2})\right)-f\left(g(x-\frac{h}{2})\right)}{(x+\frac{h}{2})-(x-\frac{h}{2})}$
$=\frac{f(M_{x}g\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})-f(M_{x}g\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{h}$
$=\frac{f(M_{x}g\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})-f(M_{x}g\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{h}\frac{\left(\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}\right)}{\left(\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}\right)}$
$=\frac{f(M_{x}g\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})-f(M_{x}g\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$
$=\frac{f(M_{x}g\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})-f(M_{x}g\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{(M_{x}g\left(x\right)+\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})-(M_{x}g\left(x\right)-\frac{h}{2}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$
$=\frac{{\delta }_{(h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}f\left(M_{x}g\left(x\right)\right)}{{\delta }_{(h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{M}_{x}g\left(x\right)}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$

$\text{よって}$
$\frac{{\delta }_{h}f\left(g\left(x\right)\right)}{{\delta }_{h}x}=\frac{{\delta }_{(h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}f\left(M_{x}g\left(x\right)\right)}{{\delta }_{(h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{M}_{x}g\left(x\right)}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$

$\frac{{∆}_{h}f\left(g\left(x\right)\right)}{{∆}_{h}x}=\frac{{∆}_{(h\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x})}f\left(g\left(x\right)\right)}{{∆}_{(h\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x})}g\left(x\right)}\frac{{∆}_{h}g\left(x\right)}{{∆}_{h}x}$
$\frac{{\nabla }_{h}f\left(g\left(x\right)\right)}{{\nabla }_{h}x}=\frac{{\nabla }_{(h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x})}f\left(g\left(x\right)\right)}{{\nabla }_{(h\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x})}g\left(x\right)}\frac{{\nabla }_{h}g\left(x\right)}{{\nabla }_{h}x}$
$\frac{{\delta }_{h}f\left(g\left(x\right)\right)}{{\delta }_{h}x}=\frac{{\delta }_{(h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}f\left(M_{x}g\left(x\right)\right)}{{\delta }_{(h\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x})}{M}_{x}g\left(x\right)}\frac{{\delta }_{h}g\left(x\right)}{{\delta }_{h}x}$
$\text{ただし}{M}_{x}g\left(x\right)=\frac{g\left(x+\frac{h}{2}\right)+g\left(x-\frac{h}{2}\right)}{2}$